Successive numbers starting from #1# are written in a rectangular grid, starting in the top left corner and snaking down diagonally to the right as shown. In which row and column does #2008# occur?

#color(white)(0)1color(white)(00)2color(white)(00)6color(white)(00)7color(white)(0)15color(white)(0)16#
#color(white)(0)3color(white)(00)5color(white)(00)8color(white)(0)14color(white)(0)"etc"#
#color(white)(0)4color(white)(00)9color(white)(0)13#
#10color(white)(0)12#
#11#

1 Answer
Feb 6, 2017

Answer:

The #9#th row, #55#th column.

Explanation:

The last number added to the #n#th (reverse) diagonal is the #n#th triangular number #T_n = 1/2n(n+1)#.

What is the smallest triangular number greater than or equal to #2008#?

Given:

#1/2n(n+1) = T_n >= 2008#

Multiply both ends by #2# to get:

#n(n+1) >= 4016#

So the value of #n# we are looking for is roughly #sqrt(4016) ~~ 63# and we find:

#T_63 = 1/2*63*64 = 2016#

#T_62 = 1/2*62*63 = 1953#

So #2008# lies on the reverse diagonal which has #T_63 = 2016# at one end.

Note that due to the boustrophedonic (like an ox ploughing a field) way in which the numbers are written, #T_63# will be written on the top row as the #63#rd term of that row.

#2008# will occur #8# rows below and to the left of it, i.e. on the #9#th row in the #55#th column.